# 165. Compare Version Numbers

Compare two version numbers version1 and version2. If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.

The . character does not represent a decimal point and is used to separate number sequences.

For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

  1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
  2. Version strings do not start or end with dots, and they will not be two consecutive dots.

# Solution

Approach 1: Scan the two strings and compare.

# Code (Python)

Approach 1:

# Code (C++)

Approach 1:

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int i1 = 0;
        int i2 = 0;
        int num1 = 0;
        int num2 = 0;
        while (i1 <= version1.size() || i2 <= version2.size())
        {
            if (i1 < version1.size() && version1[i1] != '.')
            {
                num1 = num1 * 10 + (version1[i1] - '0');
                i1++;
            }
            else if (i2 < version2.size() && version2[i2] != '.')
            {
                num2 = num2 * 10 + (version2[i2] - '0');
                i2++;
            }
            else
            {
                if (num1 > num2)
                    return 1;
                else if (num1 < num2)
                    return -1;
                num1 = 0;
                num2 = 0;
                i1++;
                i2++;
            }
        }
        return 0;
    }
};